$f(x) = \dfrac{ 9 }{ \sqrt{ 2 - \lvert x \rvert } }$ What is the domain of the real-valued function $f(x)$ ?
Answer: First, we need to consider that $f(x)$ is undefined anywhere where the radicand (the expression under the radical) is less than zero. So we know that $2 - \lvert x \rvert \geq 0$ This means $\lvert x \rvert \leq 2$ , which means $-2 \leq x \leq 2$ Next, we need to consider that $f(x)$ is also undefined anywhere where the denominator is zero. So we know that $\sqrt{ 2 - \lvert x \rvert } \neq 0$ , so $\lvert x \rvert \neq 2$ This means that $x \neq 2$ and $x \neq -2$ So we have four restrictions: $x \geq -2$ $x \leq 2$ $x \neq -2$ , and $x \neq 2$ Combining these four, we know that $x > -2$ and $x < 2$ ; alternatively, that $-2 < x < 2$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid -2< x <2\, \}$.